State dependent choice

Abstract

We propose a theory of choices that are influenced by the psychological state of the agent. The central hypothesis is that the psychological state controls the urgency of the attributes sought by the decision maker in the available alternatives. While state dependent choice is less restricted than rational choice, our model does have empirical content, expressed by simple ‘revealed preference’ type of constraints on observable choice data. We demonstrate the applicability of simple versions of the framework to economic contexts. We show in particular that it can explain widely researched anomalies in the labour supply of taxi drivers.

Appendices

Appendices

1.1 Proofs

Proof of Proposition 2

We show the following: let $c$ and $d$ be two choice functions on $\Sigma$ that have, respectively the sd-cheklists $\left( \Gamma ,\left\{ <_{A}\right\} _{A\in \Sigma }\right)$ and $\left( \Gamma ,\left\{ <_{A}^{\prime }\right\} _{A\in \Sigma }\right)$; then $c=d$.

Suppose that $c\left( A\right) \ne d\left( A\right)$ for some $A\in \Sigma$ and in particular let (possibly relabeling the choice functions) $x\in c\left( A\right)$ and $x\notin d\left( A\right)$. The latter implies that there exist $y\in A$ and $P\in \Gamma$ such that $x\notin P$ and $y\in P$. For $x\in c\left( A\right)$ it must then be the case that there exists $Q<_{A}P$ and $z\in A$ such that $y\notin Q$ and $z\in Q$. If $P\subset Q$ this is incompatible with $y\in P$, and if $Q\subset P$ then $x\notin Q$. Therefore $x\notin S_{A}(Q,<_{A})$ and $x\notin c\left( A\right)$, a contradiction.

So any sequence of the properties in the mindset $\Gamma$ generates the same behaviour and by Proposition 1 the behaviour generated by any particular sequence maximises a weak order, as claimed. $\square$

Proof of Proposition 3

Suppose that $c$ satisfies Togetherness. Define a relation $\approx$ on $X$ by $x\approx y$ iff there is no $A\in \Sigma$ such that $x\in c\left( A\right)$ and $y\in A\backslash c\left( A\right)$ or $y\in c\left( A\right)$ and $x\in A\backslash c\left( A\right)$. The relation $\approx$ is obviously reflexive and symmetric. To see that it is also transitive, suppose that $x\approx y\approx z$ and that $x\in c\left( A\right)$ and $z\in A$ for some $A\in \Sigma$. We show that $z\in c\left( A\right)$.

Since $x\approx y$ we have $x\in c\left( \left\{ x,y,z\right\} \right)$ if and only if $y\in c\left( \left\{ x,y,z\right\} \right) (\left\{ x,y,z\right\}$ is in the domain by assumption), and similarly $y\approx z$ implies that $y\in c\left( \left\{ x,y,z\right\} \right)$ if and only if $z\in c\left( \left\{ x,y,z\right\} \right)$. Therefore if $x\in c\left( \left\{ x,y,z\right\} \right)$ then $c\left( \left\{ x,y,z\right\} \right) =\left\{ x,y,z\right\}$. Therefore by Togetherness $x\in c\left( A\right)$ and $z\in A$ imply $z\in A$. If instead $x\notin c\left( \left\{ x,y,z\right\} \right)$, then $c\left( \left\{ x,y,z\right\} \right) =\varnothing$, a contradiction. $\approx$ is therefore an equivalence relation and it partitions the set of alternatives into equivalence classes, which we denote $\left[ x\right] =\left\{ y\in X:y\approx x\right\}$.

Given $A\in \Sigma$, take any $x\in c\left( A\right)$ and let $P_{A}=\left[ x\right]$. Note that $P_{A}$ is uniquely defined, and let the mindset be $\Gamma =\left\{ P_{A}:A\in \Sigma \right\}$. Since $\approx$ is an equivalence, we have $P_{A}\cap P_{B}=\varnothing$ for all distinct menus $A,B\in \Sigma$. Let the state $<_{A}$ be any linear order for which $P_{A}<_{A}P$ for all $P\in \Gamma \backslash P_{A}$. Then $A\cap P_{A}=c\left( A\right)$ (for any $y\notin c\left( A\right)$ and $x\in P_{A}$ it cannot be $y\approx x$ by the definitions of $\approx$ and $P_{A} )$. And since for all $P\in \Gamma \backslash P_{A}$ we have $P_{A}\cap P=\varnothing$, it follows that, for all $P\in \Gamma , S_{A}\left( P,<_{A}\right) =c\left( A\right)$.

Conversely, let $\left\{ \Gamma ,\left\{ <_{A}\right\} _{A\in \Sigma }\right\}$ be a menu checklist for $c$. Suppose $x,y\in c\left( A\right)$ for some $A\in \Sigma$, and suppose by contradiction that, for some $B\in \Sigma , y\in c\left( B\right)$ and $x\in B\backslash c\left( B\right)$. Then there exists $P\in \Gamma$ such that $y\in P$ and $x\notin P$. By definition of having a checklist there exists $Q\in \Gamma$ such that $S_{A}\left( Q\right) =S_{A}\left( R\right)$ for all $R\in \Gamma$ with $Q<_{A}R$. This cannot be true if $Q<_{A}P$. On the other hand, if $P<_{A}Q$ it cannot be $x,y\in c\left( A\right)$, a contradiction. $\square$

Proof of Proposition 4:

If $x\in \gamma \left( B,<_{m}\right)$ then $x$ must be in any $S_{B}(P,<_{m})$, and therefore in any $S_{A}(P,<_{m})$, so that $x\in \gamma \left( A,<_{m}\right)$ whenever $x\in A$: so $\gamma \left( B,<_{m}\right) \cap A\subseteq \gamma \left( A,<_{m}\right)$. And if $x\notin \gamma \left( B,<_{m}\right)$ and $\gamma \left( B,<_{m}\right) \cap A\ne \varnothing$, then there exists $y\in \gamma \left( B,<_{m}\right) \cap A$ that has a property which $x$ does not have. So there is $P$ for which $x\notin S_{A}(P,<_{m})$, and consequently $x\notin \gamma \left( A,<_{m}\right)$. This shows that $\gamma \left( A,<_{m}\right) \subseteq \gamma \left( B,<_{m}\right) \cap A$, and we conclude that $\gamma \left( A,<_{m}\right) =\gamma \left( B,<_{m}\right) \cap A$. $\square$

Proof of Claim 1

Let $\Gamma =\left\{ P,Q,R\right\} , M=\left\{ <_{m},<_{n}\right\} , P<_{m}Q<_{m}R$ and $R<_{n}Q<_{n}P$. Suppose $P=\left\{ x,z\right\} , Q=\left\{ y,z\right\}$ and $R=\left\{ x,y\right\}$. Then

$$\begin{aligned} \begin{array}{c} \gamma \left( \left\{ x,y\right\} ,<_{m}\right) =\left\{ x\right\} \\ \gamma \left( \left\{ x,z\right\} ,<_{m}\right) =\left\{ z\right\} \end{array} \quad \begin{array}{c} \gamma \left( \left\{ x,y\right\} ,<_{n}\right) =\left\{ y\right\} \\ \gamma \left( \left\{ x,z\right\} ,<_{n}\right) =\left\{ x\right\} \end{array} \end{aligned}$$

and therefore

$$\begin{aligned} \begin{array}{c} c\left( \left\{ x,y\right\} \right) =\left\{ x,y\right\} \\ c\left( \left\{ x,z\right\} \right) =\left\{ x,z\right\} \end{array} \quad \text {But}\quad \begin{array}{c} \gamma \left( \left\{ x,y,z\right\} ,<_{m}\right) =\left\{ z\right\} \\ \gamma \left( \left\{ x,y,z\right\} ,<_{n}\right) =\left\{ y\right\} \end{array} \end{aligned}$$

so that $c\left( \left\{ x,y,z\right\} \right) =\left\{ y,z\right\}$. We conclude that $c$ violates both Expansion and Property $\beta$. $\square$

Proof of Proposition 6

We will find it convenient to write sd-WARP in an equivalent way:²¹

sd-WARP (restated): $c\left( A\right) \subseteq B\Rightarrow c\left( B\right) \cap A\subseteq c\left( A\right)$.²²

Necessity. As a preliminary, we say that ‘$x m-$tops $y$’, written $xT_{m}y$, if there is a state $<_{m}$ and a property $P_{i}$ such that $x\in P_{i}, y\notin P_{i}$ and $y\in P_{j}\Rightarrow x\in P_{j}$ for all $P_{j}$ such that $P_{j}<_{m}P_{i}$. Observe that for all $D\in \Sigma$ and $<_{m}\in M, x\in \gamma \left( D,<_{m}\right)$ only if there is no $y\in D$ such that $yT_{m}x$.

Let $A,B\in \Sigma$ be such that $c\left( A\right) \subseteq B$. The statement of the proposition is trivially true if $A\cap \gamma \left( B,<_{m}\right) =\varnothing$ for all $<_{m}\in M$ (in which case $A\cap c\left( B\right) =\varnothing$), so suppose that $A\cap \gamma \left( B,<_{m}\right) \ne \varnothing$ for some $<_{m}\in M$. Then

$$\begin{aligned} A\cap c\left( B\right)= & {} A\cap \bigcup _{<_{m}\in M}\gamma \left( B,<_{m}\right) \\= & {} \bigcup _{<_{m}\in M}A\cap \gamma \left( B,<_{m}\right) \subseteq \bigcup _{<_{m}\in M}\gamma \left( A,<_{m}\right) \\= & {} c\left( A\right) \end{aligned}$$

where the inclusion is proved with the following reasoning. Since $c\left( A\right) \subseteq B$, for all $<_{m}\in M$ we have $\gamma \left( A,<_{m}\right) \subseteq B$. So in particular there is no $y\in A\backslash B$ that $m-$tops any $x\in \gamma \left( A,<_{m}\right)$. Therefore for all $x\in \gamma \left( B,<_{m}\right) \cap A$ we also have $x\in \gamma \left( A,<_{m}\right)$ (if not, there would exist $y \in A\backslash B$ with $yT_{m}x$). We conclude that, for all $<_{m}\in M, \gamma \left( B,<_{m}\right) \cap A\subseteq \gamma \left( A,<_{m}\right)$, from which the desired inclusion follows.

Sufficiency. Let sd-WARP hold. We construct an environmental checklist explicitly, then show that it retrieves $c\left( A\right)$ for each menu $A\in \Sigma$. Let $\Gamma =\left\{ \left\{ x\right\} _{x\in X}\right\}$, and let $\left| X\right| =n$.

An $a-$path is a sequence $a=\left\{ x_{i}\right\} _{i=1,...n}$ of distinct alternatives $x_{1},x_{2},...x_{n}$ defined recursively as follows. $x_{1}\in c\left( X\right)$ and, for all $\ i>1, x_{i}\in c\left( X\backslash \left\{ x_{1},...,x_{i-1}\right\} \right)$. Denote by $\mathcal {\alpha }$ the collection of $a-$paths, and note that each $a-$path covers all of the alternatives in $X$. Construct $M$ by setting, for each $a\in \alpha$:

$$\begin{aligned} \left\{ x_{i}\right\} <_{a}\left\{ x_{j}\right\} \text { if and only if } i<j\text { and }x_{i},x_{j}\in a \end{aligned}$$

We now show that this construction retrieves choice.

Fix an arbitrary menu $A\in \Sigma$, let $x\in c\left( A\right)$, and suppose by contradiction that for each $<_{m}\in M$, there is an alternative $w$ such that $\left\{ w\right\} <_{m}\left\{ x\right\}$. For each $<_{m}$ let $\left\{ y_{m}\right\}$ denote the $<_{m}-$maximal property in $A$, that is $\left\{ y_{m}\right\} <_{m}\left\{ z\right\}$ for all $z\in A\backslash \left\{ y_{m}\right\}$. By construction we have that $y_{m}\in \gamma \left( A_{m},<_{m}\right)$ where $A_{m}=\left\{ y_{m}\right\} \cup \left\{ z\in X:\left\{ y_{m}\right\} <_{m}\left\{ z\right\} \right\}$. Observe that by assumption there is no $<_{m}$ such that $x\in \gamma \left( A_{m},<_{m}\right)$ (otherwise $\left\{ x\right\}$ would be maximal in $A$ for some state). Moreover, by construction it must also be that $A\subseteq A_{m}$, for otherwise it would not be true that $\left\{ y_{m}\right\} <_{m}\left\{ z\right\}$ for all $z\in A\backslash \left\{ y_{m}\right\}$. If for any of the $A_{m}$ it is the case that $c\left( A_{m}\right) \subseteq A$, then by sd-WARP it would follow that $c\left( A\right) \cap A_{m}\subseteq c\left( A_{m}\right)$, contradicting $x\notin \gamma \left( A_{m},<_{m}\right)$. So suppose not, so that $c\left( A_{m}\right) \backslash A\ne \varnothing$, and consider $A_{m1}=A_{m}\backslash \left\{ z_{1}\right\}$ where $z_{1}\in c\left( A_{m}\right) \backslash A$. As before, either $c\left( A_{m1}\right) \subseteq A$, so that the contradiction $c\left( A\right) \cap A_{m1}\subseteq c\left( A_{m1}\right)$ follows; or $c\left( A_{m1}\right) \backslash A\ne \varnothing$. More in general, proceed recursively setting $A_{mj}=A_{mj-1}\backslash \left\{ z_{j}\right\}$ where $z_{j}\in c\left( A_{mj}\right) \backslash A$ whenever $c\left( A_{mj}\right) \backslash A\ne \varnothing$ and $j>1$. At each step either $c\left( A_{mj}\right) \subseteq A$, implying $c\left( A\right) \cap A_{mj}\subseteq c\left( A_{mj}\right)$; or $c\left( A_{mj}\right) \backslash A\ne \varnothing$. Since $X$ is finite there exists a $j^{*}$ such that $c\left( A_{mj^{*}}\right) \backslash A=\varnothing$, generating the desired contradiction.

Suppose now that $x\in A\backslash c\left( A\right)$, and that in contradiction there exists some state $<_{m}\in M$ such that $x\in \gamma \left( A,<_{m}\right)$. By construction it must be that $x\in c\left( B\right)$ where $B=\left\{ x\right\} \cup \left\{ y\in X:\left\{ x\right\} <_{m}\left\{ y\right\} \right\}$, and that $A\subseteq B$. If $A=B$ we have an immediate contradiction. Otherwise, then $c\left( A\right) \subset B$, so that by sd-WARP $c\left( B\right) \cap A\subseteq c\left( A\right)$ also follows, implying $x\in c\left( A\right)$, a contradiction. $\square$

Proof of Claim 2

By example: $\square$

Example 4

(Menu driven state dependent but not indecisive) $c\left( \left\{ x,y,z\right\} \right) =\left\{ x,y\right\} , c\left( \left\{ x,y\right\} \right) =\left\{ x,y\right\} , c\left( \left\{ x,z\right\} \right) =\left\{ x\right\} , c\left( \left\{ y,z\right\} \right) =\left\{ z\right\}$.

Example 5

(Environment driven state dependent but not indecisive) $c\left( \left\{ x,y,z\right\} \right) =\left\{ x,y\right\} , c\left( \left\{ x,y\right\} \right) =\left\{ x,y\right\} , c\left( \left\{ x,z\right\} \right) =\left\{ x,z\right\} , c\left( \left\{ y,z\right\} \right) =\left\{ y,z\right\}$.

Example 6

(Indecisive but neither menu driven state dependent nor environment driven state dependent) $c\left( \left\{ x,y,z\right\} \right) =\left\{ y\right\} , c\left( \left\{ x,y\right\} \right) =\left\{ x,y\right\} , c\left( \left\{ x,z\right\} \right) =\left\{ z\right\} , c\left( \left\{ y,z\right\} \right) =\left\{ y\right\}$.

1.2 Examples

1.2.1 Property $\alpha$ is not sufficient for environment driven state dependent choice

Suppose $c\left( \left\{ x,y,z\right\} \right) =\left\{ x\right\}$ and $c\left( \left\{ x,y,w\right\} \right) =\left\{ x,y\right\}$. Although these two choices do not violate Property $\alpha$, it is not possible to find a mindset $\Gamma$, a set of states $M$ and a choice function $\gamma$ such that

$$\begin{aligned} c\left( A\right) =\bigcup _{<_{m}\in M}\gamma \left( A,<_{m}\right) \end{aligned}$$

To see this, suppose to the contrary that $y\in \gamma \left( \left\{ x,y,w\right\} ,<_{m}\right)$ for some $m\in M$. Since $y\notin c\left( \left\{ x,y,z\right\} \right)$, it must be that there is an alternative $i\in \left\{ x,y,z\right\}$ such that $i\in P_{i}, y\notin P_{i}$ and $y\in P_{j}\Rightarrow i\in P_{j}$ for all $P_{j}$ such that $P_{j}<_{m}P_{i}$. If $i=x$, then it could not be that $y\in \gamma \left( \left\{ x,y,w\right\} ,<_{m}\right)$; while if $i=z$, then either $x\notin \gamma \left( \left\{ x,y,w\right\} ,<_{m}\right)$, or it must be that $z\in c\left( \left\{ x,y,z\right\} \right)$. In either case we have a contradiction.

1.2.2 sd-WARP implies Property $\alpha$

Let sd-WARP hold, and suppose that there are sets $A$ and $B$ such that $A\subset B$ but that in contradiction to Property $\alpha$ there is some $x\in C\left( B\right) \cap A$ such that $x\notin C\left( A\right)$. Since $A\subset B$ it also follows that $C\left( A\right) \subset B$, which together with $x\in A\backslash C\left( A\right)$ and sd-WARP implies $x\notin c\left( B\right)$, contradiction.

1.2.3 Independence of the two state dependent choice models

The models we have considered in the paper (menu and environment driven states) are logically independent. We illustrate this with a simple example.²³ Let $A=\left\{ h_{s},h_{L},u_{s}\right\}$ and $B=A\cup \left\{ u_{L}\right\}$, where $h_{i}$ and $u_{i}$ stand for healthy and unhealthy food items, with the index 1 denoting a smaller portion than Index 2. Let $c_{1}\left( A\right) =c_{2}\left( A\right) =\left\{ h_{s},h_{L}\right\} , c_{1}\left( B\right) =\left\{ u_{s},u_{L}\right\}$ and $c_{2}\left( B\right) =\left\{ h_{L},u_{L}\right\}$. Then $c_{s}$ is menu driven but not environment driven state dependent (Togetherness holds and sd-WARP fails), while $c_{2}$ is environment driven but not menu driven state dependent (sd-WARP holds and Togetherness fails). The states for $c_{s}$ could be e.g. $\left\{ h_{s},h_{L}\right\} <_{A}\left\{ u_{s},u_{L}\right\}$ and $\left\{ u_{s},u_{L}\right\} <_{B}\left\{ h_{s},h_{L}\right\}$, in line with the explanation that the decision maker can stick to healthy food when they are the majority, but adding an extra unhealthy food item switches the state to ‘gluttony’. And for $c_{2}$ we could have $\left\{ h_{s},h_{L}\right\} <_{c}\left\{ h_{L},u_{L}\right\} <_{c}\left\{ u_{L}\right\}$ when the decision maker is in a cool state, and sticks to his diet; while in a depressed state he seeks satisfaction in large portion sizes, e.g. $\left\{ u_{L}\right\} <_{d}\left\{ h_{s},h_{L}\right\} <_{d}\left\{ h_{L},u_{L}\right\}$.

1.3 An alternative characterisation of menu driven state dependent choice

A slightly different angle on the behavioural restriction of menu driven state dependent choice is the following:

All or Nothing: If the choices from two different menus overlap, then the choice from one menu consists of those available alternatives that are chosen from the other menu. Formally, for all $A,B\in \Sigma : c\left( A\right) \cap c\left( B\right) \ne \varnothing \Rightarrow c\left( A\right) \cap c\left( B\right) =c\left( A\right) \cap B$.

All or Nothing describes either a form of ‘behavioural discontinuity’ or of ‘behavioural inertia’, excluding other possibilities. That is, when moving from a menu $A$ to a different menu $B$, either the agent’s behaviour changes abruptly (no alternative is chosen from both menus) or whatever was originally chosen in $A$ and is still available in $B$, it remains chosen, and no new alternatives are added to the choice.²⁴

Corollary 1

A choice function is a menu driven state dependent choice if and only if it satisfies All or Nothing.

Proof

We show the equivalence of Togetherness and All or Nothing. Suppose that Togetherness holds and that $c\left( A\right) \cap c\left( B\right) \ne \varnothing$. Obviously for any $x\in c\left( A\right) \cap c\left( B\right) \$we have $x\in c\left( A\right) \cap B$, that is $c\left( A\right) \cap c\left( B\right) \subseteq c\left( A\right) \cap B$. For the converse inclusion, for any $x\in c\left( A\right) \cap B$ either $c\left( A\right) \cap c\left( B\right) =\left\{ x\right\}$ or there exists $y\ne x$ with $y\in c\left( A\right) \cap c\left( B\right) \$ and so by Togetherness $x\in c\left( B\right)$ (otherwise, $x\in B\backslash c\left( B\right)$ would violate Togetherness). This shows that $c\left( A\right) \cap B\subseteq c\left( A\right) \cap c\left( B\right)$ and we conclude that $c\left( A\right) \cap c\left( B\right) =c\left( A\right) \cap B$.

Conversely, suppose that Togetherness is violated, that is there exist $A,B\in \Sigma$ and $x\in A\backslash c\left( A\right) , y\in c\left( A\right) , y\in c\left( B\right)$ but $x\in c\left( B\right)$. Then $c\left( A\right) \cap c\left( B\right) \ne \varnothing$. Moreover, $x\notin c\left( A\right) \cap c\left( B\right)$ while $x\in c\left( B\right) \cap A$, so that $c\left( A\right) \cap c\left( B\right) \ne c\left( B\right) \cap A$, violating All or Nothing.

1.4 A technical remark on the relation of Proposition 6 with a theorem by Litvakov

We establish a connection with a not very well-known result by Litvakov (1981), asserting that if a choice function $c$ satisfies Property $\alpha$ and Chernoff’s Postulate 5 (if $c(B)\subseteq A\subseteq B$ then $c(A)=c(B)$ ),then it can be expressed as the union of choice correspondences $c_{i}$ (i.e. $c\left( A\right) =\cup _{i}c\left( A\right)$ for all $A$), each of which satisfies IIA. Because Litvakov (1981) is written in Russian, a more useful general reference is Aizerman and Aleskerov (1995), Theorem 5.6(b) (and Theorem 5.2.1 for the other direction).

The following result can be proved (details available from the authors):

Claim 3

A choice function $c$ satisfies sd-WARP if and only if it satisfies Property $\alpha$ and Chernoff’s Postulate 5.

Then, it can be seen that Proposition 6 can also be derived indirectly as the combined consequence of Litvakov’s theorem, the above claim, and Proposition 1 in this paper.